use the shell method to compute the volume obtained by rotating the region enclosed by the graphs as…

Explanation:

Step1: Recall the Shell Method formula

The Shell Method formula for rotating about the (y -)axis is (V = 2\pi\int_{a}^{b}x\left(f(x)-g(x)\right)dx), where (f(x)) is the upper - function and (g(x)) is the lower - function, and ([a,b]) is the interval of integration.

Here, (f(x)=2-(x^{2}+1)^{-2}), (g(x)=(x^{2}+1)^{-2}), and (a = 0), (b = 6). Then (f(x)-g(x)=2 - 2(x^{2}+1)^{-2}).

Step2: Set up the integral

So, (V=2\pi\int_{0}^{6}x\left(2 - 2(x^{2}+1)^{-2}\right)dx=4\pi\int_{0}^{6}\left(x-\frac{x}{(x^{2}+1)^{2}}\right)dx)

Step3: Integrate term - by - term

• For the first integral (\int xdx=\frac{x^{2}}{2}+C_1).
• For the second integral, use substitution. Let (u = x^{2}+1), then (du = 2xdx) and (xdx=\frac{1}{2}du). So (\int\frac{x}{(x^{2}+1)^{2}}dx=\frac{1}{2}\int u^{-2}du). Using the power rule (\int u^{n}du=\frac{u^{n + 1}}{n+1}+C) ((n\neq - 1)), we have (\frac{1}{2}\int u^{-2}du=\frac{1}{2}\times\frac{u^{-1}}{-1}+C=-\frac{1}{2(x^{2}+1)}+C_2)

Step4: Evaluate the definite integral

[ \begin{align*} V&=4\pi\left[\frac{x^{2}}{2}+\frac{1}{2(x^{2}+1)}\right]_{0}^{6}\ &=4\pi\left(\left(\frac{6^{2}}{2}+\frac{1}{2(6^{2}+1)}\right)-\left(\frac{0^{2}}{2}+\frac{1}{2(0^{2}+1)}\right)\right)\ &=4\pi\left(\frac{36}{2}+\frac{1}{2\times37}- \frac{1}{2}\right)\ &=4\pi\left(18+\frac{1}{74}-\frac{1}{2}\right)\ &=4\pi\left(\frac{18\times74 + 1-37}{74}\right)\ &=4\pi\left(\frac{1332 + 1-37}{74}\right)\ &=4\pi\left(\frac{1296}{74}\right)\ &=4\pi\times\frac{648}{37}\ &=\frac{2592\pi}{37} \end{align*} ]

Answer:

(\frac{2592\pi}{37})